Integrate. $ \int \csc(x)\cot(x)\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $-\cot x + C$ (Choice B) B $-\tan x + C$ (Choice C) C $-\sec x + C$ (Choice D) D $-\csc x + C$
Solution: We need a function whose derivative is $\csc(x)\cot(x)$. We know that the derivative of $\csc(x)$ is $-\csc(x)\cot(x)$, so let's start there: $\dfrac{d}{dx} \csc(x) = -\csc(x)\cot(x)$ Now let's multiply by $-1$ : $\dfrac{d}{dx}\left[ -1\csc(x)\right] = -1\dfrac{d}{dx}\csc(x) =\csc(x)\cot(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int \csc(x)\cot(x)\,dx =- \csc(x)\, + C$ The answer: $- \csc(x)\, + C$